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Three Dimensional Geometry Set-5 MCQ Online Practice Test | Class 12 | Quiz Tweek

Three Dimensional Geometry Set-5

Practice Important MCQs for Three Dimensional Geometry — Mathematics, Class 12.

Three Dimensional Geometry is central in Class 12 and competitive exams because it builds the ability to analyze lines, planes, distances, and angles in 3D using vectors and coordinate methods. Mastery of this chapter strengthens problem-solving in board exams and is heavily used in JEE/NEET for questions on shortest distances, intersection/containment conditions, and plane-line geometry—often with efficient vector cross/dot product techniques.

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Q1. Find the perpendicular distance from the point $P(2,-1,3)$ to the plane $2x-3y+6z+1=0$ .

(A)

$\dfrac{13}{7}$

(B)

$3$

(C)

$\dfrac{26}{7}$

(D)

$\dfrac{\sqrt{26}}{7}$

Q2. Find the equation of the plane which passes through the point $A(1,2,-1)$ and contains the line of intersection of the planes $x+2y+z-3=0$ and $2x-y+3z+1=0$ .

(A)

$4x+3y+5z-5=0$

(B)

$3x+4y+5z-5=0$

(C)

$4x+3y+5z+5=0$

(D)

$4x+5y+3z-5=0$

Q3. The plane is $x+2y+2z+3=0$ and the line $L$ is given by $\mathbf{r}=(1,0,1)+t(2,-1,0)$ . Find the shortest distance from $L$ to the plane and the coordinates of the point on the plane nearest to $L$ .

(A)

Distance $=2$ , nearest point $\left(\tfrac{1}{3},\tfrac{4}{3},\tfrac{1}{3}\right)$

(B)

Distance $=\dfrac{2}{3}$ , nearest point $\left(\tfrac{1}{3},-\tfrac{4}{3},-\tfrac{1}{3}\right)$

(C)

Distance $=2$ , nearest point $\left(\tfrac{1}{3},-\tfrac{4}{3},\tfrac{1}{3}\right)$

(D)

Distance $=2$ , nearest point $\left(\tfrac{1}{3},-\tfrac{4}{3},-\tfrac{1}{3}\right)$

Q4. Find the shortest distance between the skew lines $L_1:\mathbf{r}=(1,0,0)+t(1,1,1)$ and $L_2:\mathbf{r}=(0,1,2)+s(1,2,3)$ . Also find the points $P$ on $L_1$ and $Q$ on $L_2$ at which this shortest distance is attained.

(A)

$\text{Distance}=\dfrac{1}{\sqrt{6}},\ P\!\left(\tfrac{4}{3},\tfrac{7}{3},\tfrac{7}{3}\right),\ Q\!\left(\tfrac{3}{2},2,\tfrac{5}{2}\right)$

(B)

$\text{Distance}=\dfrac{1}{\sqrt{6}},\ P\!\left(-\tfrac{4}{3},-\tfrac{7}{3},-\tfrac{7}{3}\right),\ Q\!\left(-\tfrac{3}{2},-2,-\tfrac{5}{2}\right)$

(C)

$\text{Distance}=\dfrac{1}{\sqrt{6}},\ P\!\left(-\tfrac{4}{3},-\tfrac{7}{3},-\tfrac{7}{3}\right),\ Q\!\left(\tfrac{3}{2},2,\tfrac{5}{2}\right)$

(D)

$\text{Distance}=\dfrac{1}{2\sqrt{6}},\ P\!\left(-\tfrac{1}{3},-\tfrac{2}{3},-\tfrac{1}{3}\right),\ Q\!\left(-\tfrac{3}{4},-1,-\tfrac{5}{4}\right)$

Q5. A variable plane with intercepts $a,b,c\neq0$ on the coordinate axes has equation $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ and passes through the fixed point $(1,2,3)$ . The triangle formed by its intercepts has centroid $G(x,y,z)$ . The locus of $G$ is:

(A)

$\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{3}{z}=3$

(B)

$\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1$

(C)

$\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{3}{z}=1$

(D)

$x+2y+3z=3$

Q6. Find the perpendicular distance from the point $P(1,2,-1)$ to the plane $2x-3y+6z-4=0$ .

(A)

$1$

(B)

$2$

(C)

$3$

(D)

$\sqrt{5}$

Q7. For which values of $k$ does the line

\begin{aligned} x &= 1+t \\ y &= 2+kt \\ z &= 3+2t \end{aligned}

touch the sphere $x^2+y^2+z^2-2x+4y-6z+3=0$ ?

(A)

$k=0$

(B)

$k=\pm\frac{5}{11}$

(C)

$k=\pm\sqrt{\frac{5}{11}}$

(D)

$k=\pm\frac{5}{\sqrt{11}}$

Q8. Find the equation of the plane passing through the point $A(1,0,2)$ , parallel to the line

\begin{aligned} x &= 3+t \\ y &= 1+2t \\ z &= -1+t \end{aligned}

and perpendicular to the plane $2x-y+z-3=0$ .

(A)

$3x+y-5z+7=0$

(B)

$3x+y-5z-7=0$

(C)

$-3x-y+5z-7=0$

(D)

$2x-y+z-3=0$

Q9. Assertion (A): If a line $l$ is perpendicular to a plane $\Pi$ , then $l$ is perpendicular to every line lying in $\Pi$ .

Reason (R): A line is perpendicular to a plane if it is perpendicular to two distinct lines in the plane that intersect at the point of intersection with the given line.

(A)

Both A and R are true and R is a correct explanation of A.

(B)

Both A and R are true but R is not a correct explanation of A.

(C)

A is true but R is false.

(D)

A is false but R is true.

Q10. Let

\begin{aligned} L_1: x &= 1+2\lambda \\ y &= 2-\lambda \\ z &= 3+\lambda \end{aligned} \quad\text{and}\quad \begin{aligned} L_2: x &= 3+\mu \\ y &= 0+2\mu \\ z &= 1 \end{aligned}.

Points $P\in L_1$ and $Q\in L_2$ are the pair for which segment $PQ$ is the shortest joining the two skew lines. Find $P,\;Q$ and the length $PQ$ .

(A)

$P=\left(\frac{7}{3},\frac{4}{3},\frac{11}{3}\right),\; Q=\left(\frac{17}{5},\frac{4}{5},1\right),\; PQ=\dfrac{4\sqrt{30}}{15}$

(B)

$P=\left(\frac{5}{2},\frac{3}{2},\frac{7}{2}\right),\; Q=\left(\frac{17}{5},\frac{4}{5},1\right),\; PQ=\dfrac{16}{\sqrt{30}}$

(C)

$P=\left(\frac{7}{3},\frac{4}{3},\frac{11}{3}\right),\; Q=\left(\frac{17}{5},\frac{4}{5},1\right),\; PQ=\dfrac{8\sqrt{30}}{15}$

(D)

$P=\left(\frac{7}{3},\frac{4}{3},\frac{11}{3}\right),\; Q=\left(\frac{14}{5},\frac{6}{5},1\right),\; PQ=\dfrac{8\sqrt{30}}{15}$

Q11. Find the shortest distance from the point $P(4,1,0)$ to the line $L:\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z-3}{3}$ .

(A)

$\dfrac{3\sqrt{19}}{7}$

(B)

$\dfrac{6}{\sqrt{7}}$

(C)

$\dfrac{3\sqrt{38}}{\sqrt{14}}$

(D)

$\dfrac{\sqrt{342}}{7}$

Q12. The skew lines $L_1$ and $L_2$ are given by $L_1:\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{-1}$ and $L_2:\dfrac{x-2}{2}=\dfrac{y+1}{-1}=\dfrac{z+1}{2}$ . The shortest distance between $L_1$ and $L_2$ is

(A)

$\dfrac{7}{\sqrt{2}}$

(B)

$\dfrac{7\sqrt{2}}{4}$

(C)

$\dfrac{5}{\sqrt{2}}$

(D)

$\dfrac{7}{2}$

Q13. Find the equation of the plane which contains the line $L:\dfrac{x-1}{1}=\dfrac{y-0}{1}=\dfrac{z-1}{1}$ and the point $B(0,2,3)$ .

(A)

$x - y + z - 2 = 0$

(B)

$x + y - z = 0$

(C)

$y + z - 1 = 0$

(D)

$z - y - 1 = 0$

Q14. Let $P_1: x+y+z-3=0$ and $P_2: 2x-y+z-1=0$ . Find the equation(s) of the plane(s) through the line of intersection of $P_1$ and $P_2$ which make an angle of $45^\circ$ with the plane $P_3: x-2y+2z-4=0$ .

(A)

$(10+3\sqrt{6})x+(4-\sqrt{6})y+(8+\sqrt{6})z-16+\sqrt{6}=0$ and $(10-3\sqrt{6})x+(4+\sqrt{6})y+(8-\sqrt{6})z-16-\sqrt{6}=0$

(B)

$(10+6\sqrt{6})x+(4-3\sqrt{6})y+(8+3\sqrt{6})z-20-3\sqrt{6}=0$ and $(10-6\sqrt{6})x+(4+3\sqrt{6})y+(8-3\sqrt{6})z-20+3\sqrt{6}=0$

(C)

$(10+6\sqrt{6})x+(4+3\sqrt{6})y+(8+3\sqrt{6})z-16+3\sqrt{6}=0$ and $(10-6\sqrt{6})x+(4-3\sqrt{6})y+(8-3\sqrt{6})z-16-3\sqrt{6}=0$

(D)

Only one such plane exists: $(10+6\sqrt{6})x+(4-3\sqrt{6})y+(8+3\sqrt{6})z-16+3\sqrt{6}=0$

Q15. Let $L_1: \mathbf{r}=(1,2,3)+s(1,-1,2)$ and $L_2: \mathbf{r}=(2,0,1)+t(2,1,-1)$ . The length of the shortest segment joining $L_1$ and $L_2$ is

(A)

$\dfrac{17\sqrt{35}}{35}$

(B)

$\dfrac{\sqrt{10115}}{70}$

(C)

$\dfrac{17\sqrt{7}}{35}$

(D)

$\dfrac{17}{5}$

...and 5 more challenging questions available in the interactive simulator.

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